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Zvi Rosen Algebraic Geometry Notes Richard Borcherds
1. Thursday, August 23, 2012
Course website: http://math.berkeley.edu/∼reb/256A
Text: Hartshorne
3 2 2 2
Example 1.1. Find solutions (x,y,z) ∈ Z to x +y = z .
2
(1) Algebraic Solution: x = (z−y)(z−y). Assume x,y,z coprime, and x odd. So (z−y),(z+y)
are both squares.
Set z −y = r2,z +y = s2, with r,s odd positive numbers.
r2 +s2 s2 −r2
z = 2 , y = 2 , x = rs.
Then, for example, taking (r,s) = (1,3) gives us (x,y,z) = (3,4,5).
(2) Geometric solution: Solve X2 + Y2 = 1 in rationals, where X = x and Y = y. Therefore,
z z
we are looking for rational points on the circle.
Finding real points is easy, just take X = sinθ,Y = cosθ, however this doesn’t help us
very much.
Instead, fix the point at (−1,0) and look at lines from that point that intersect the
tangent to the circle at (1,0). Where do these lines intersect the circle?
If the line intersects the tangent at (1,t), then it intersects the circle at (X,Y ) where
t = Y
X+1
2
⇒Y =t(X+1)⇒t2(X+1)2+X2=1⇒X=1−t ,Y = 2t .
2 2
1+t 1+t
Therefore, t rational ⇒ X,Y rational. 1
So rational points on the circle almost correspond to rational t. t = 2 means that
X=3,Y = 4. The map (X,Y)→t is a BIRATIONAL map from circle to line. Birational
5 5
means isomorphism except on a set o codimension ≥ 1.
(For smooth manifolds, “birational” maps are trivial. Any smooth manifold of dimension
n can be cut along n−1 dimensional sub manifolds so it is a union of Rns.)
Because the circle is the group of rotations, this means the set of pythagorean triples is
a group.
Product of group:
(X ,Y )×(X ,Y ) = (X X −Y Y ,X Y +X Y ).
1 1 2 2 1 2 1 2 1 2 2 1
This algebraic formula works to send any commutative ring to a group.
2 3 2
Example 1.2. Solve y = x +x in integers. One solution is (3,6).
Wecan draw the graph: INSERT GRAPH HERE.
The graph has a singularity at the origin. If we send a line from the origin, it will intersect at a
point (x,y), and the line has slope t = y.
x
Given t, determine y,x. Solving for x will give a CUBIC equation; two of the roots are 0 and
the third root will be rational.
3 2 3 2 2 2 2
y = tx ⇒ t x = x +x ⇒t =x+1⇒x=t −1,y=t(t −1).
This example gives us a taste of singularities, which we generally want to get rid of. We accom-
plish this through resolution of singularities which map a singular curve to a smooth curve above
it.
2 3 2
Here, we took the smooth curve, the line t and mapped to this singular curve y = x +x .
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Zvi Rosen Algebraic Geometry Notes Richard Borcherds
n n n n n
Example 1.3. Find rational solutions of x +y = 1 ⇔ X +Y = Z for integers, or Fermat’s
Last Theorem. This shows us that Algebraic Geometry over Q is really hard.
Example 1.4. Dudeney puzzle: x3+y3 = 9 in rationals. One solution is (1,2). Find another one.
His answer was:
415280564497 3 676702467503 3
38671682660 + 348671682660 =9.
How did he find it? Draw the curve x3 +y3 = 9. The curve has no double point. Suppose that
(x ,y ),(x ,y ) are two rational points on the curve; then, the line through them intersects the
1 1 2 2
curve in a third point. Since the sum of roots are rational, the final point of intersection is rational.
Asimilar technique would be to take the tangent to a point on the curve, and see where else it
intersects. This is the “Chord-tangent process”. This is essentially a group law.
More explicitly, the group law goes: Fix some rational point, call it the identity. Three points
lie on a straight line means that their sum is the identity.
Still it is not quite a group, because the point at infinity is missing. This is evident if you take
the line between (1,2) and (2,1); this line does not meet the curve again except at infinity.
In this case, we are working with projective varieties rather than affine varieties.
2
˚
Definition 1.5. Projective Space: “Add points at infinity.” Points of affine space A are written
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(x,y). Points of the projective space P are written (x : y : z) not all 0, with (x : y : z) = (λx : λy :
λz).
The projective plane contains the affine plane (x : y : 1), the affine line (x : 1 : 0) (at infinity),
and another point (1 : 0 : 0).
3 3 3 3 3
How is x +y =9 a curve in projective plane? Make it homogeneous: x +y = 9z – this is a
projective cubic curve. Its points form a group. Example of an abelian variety (group and also a
projective variety).
Exampleofan“abelianlinear group” (the old name for the symplectic group) that is not abelian.
Theorem 1.6 (Pappus’ Hexagon Theorem). Two lines, three points selected on each line labeled
A,B,C and a,b,c.Draw lines across except between the same letters. The resulting three points of
intersections are collinear.
Theorem 1.7 (Pascal’s Theorem). Take a conic (ellipse, parabola, hyperbola), with a similar set
up of six points. The intersection points here are also linear. Clearly, the case with two lines in
Pappus’ Theorem is a degenerate case of this theorem.
Proof. Label the lines L and fix any new point P on the ellipse. Suppose Line L is given by
i i
equation Pi = 0. Look at the equation
P1P2P3 −λP4P5P6 = 0.
Choose λ so P is a solution of this. Look at the equation X = 0, X a degree two polynomial, of
the conic. There are 7 points on the conic and cubic (the six we started with and the new P).
Theorem 1.8 (Bezout’s Theorem). Curves of degree m,n intersect in ≤ m,n points unless they
have a curve in common.
Therefore, the cubic and the conic must have a common component. So the cubic factorizes as
equation of conic times equation of line.
Sloppy Proof of Bezout’s Theorem. As stated in old books, the theorem was “Curves of degree m,n
have mn intersection points.” (False, as stated). The sloppy proof was “what is true up to the limit
is true at the limit.” (Obviously a false statement.) Given the two equations f(x,y) = 0 and
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Zvi Rosen Algebraic Geometry Notes Richard Borcherds
g(x,y) = 0, you deform both until each is a product of linear factors. Assuming that these lines
are nonparallel and distinct, they will have the desired number of intersection points.
Kakeya set in R2 is a set containing a unit line segment in every direction. Besikovich proved
that a Kakeyu set can have arbitrarily small area. Thomas Wolfe conjectured the following, later
proved by Dini.
Theorem 1.9 (Finite field Kakeya Conjecture). The size of a Kakeya set in Fn for a finite field
n
F is at least c |F| , where c = some constant not depending on F.
n n
Dini’s Proof. (1) A Kakeya set cannot lie in a hypersurface f(x ,...) = 0 of degree < |F|. If f
1
is a polynomial of degree < |F| defining a hypersruface containing a Kakeya set, then
f = f +f +....(d = highest degree component of f.)
d d−1
For all v we can find x such that f(x+vt) vanishes for all t ∈ F, so the coefficient f (v)
d
of td vanishes. As this is true for any v and deg f < |F|, we must have f = 0 so f = 0.
d d
(A polynomial of degree < |F| cannot vanish on all points of F)
(2) Space of polynomials of degree ≤ d in n variables is a vector space of dimension n+d so for
n
any set with ≤ t has many elements, can find a nonzero polynomial of degree ≤ d vanishing
on this set.
So any Kakeya set has
n+|F|−1 |F|n
≥ n ≥ n! elements.
3 3 3 3
Example 1.10 (27 Lines on a Cubic Surface). Consider the cubic surface w +x +y +z = 0.
This is a cubic surface in P3.
It contains a line (1 : −1 : t : −t). This surface has many symmetries:
(1) We can permute these four entries using S4 which has order 24.
√ 3 4
(2) We multiply any coordinate by 3−1. Gives a group of order 3 (not 3 ). (Multiplying all
coordinates by w is identity)
So we have a group of order 33 · 24 acting on this cubic surface. The line above has 27 images
under this group.
2. Tuesday, August 28, 2012
2.1. Affine Varieties.
Definition 2.1. Let k be a field, for convenience C. Affine space = kn, but we “forget” where the
origin is. What does this mean? Consider the automorphism groups.
Aut(kn) = GL (k).
n
Aut(An) = kn •GL (k).
n
In taking the affine space, we allow translations.
Definition2.2. Affinegeometry=thepropertiesofkn invariantundertheaffinegroupkn•GL (k).
n
Example 2.3. The set of conics is invariant under kn · GLn(k); however, the set of circles is not,
since a linear transformation can turn it into an ellipse.
Definition 2.4. An algebraic set in kn = An is the set of zeros of some set of polynomials.
2
Example 2.5. The parabola is an algebraic set, as the zero set of the equation y −x .
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Zvi Rosen Algebraic Geometry Notes Richard Borcherds
Definition 2.6. The Zariski topology is the topology taking algebraic sets as the closed sets. This
topology is non-Hausdorf!!
Proof. To confirm topology axioms, check that algebraic sets are closed under finite unions and
arbitrary intersections. Suppose X = zero set of {p ,p ,...}, and Y = zero set of {q ,q ,...}.
1 2 1 2
(1) X ∪Y = zero set of {piqj}.
(2) X ∩Y ∩Z ∩··· = the zero set of the union of all of the polynomials.
Remark 2.7. Topology of A1. The closed sets are:
(1) Whole space. (Zeros of the empty set)
(2) Finite subset. (Zeros of (x − a )(x − a )···).
1 2
Points are closed. (T ), but any two nonempty open sets have non-empty intersection, assuming k
1
infinite.
Remark 2.8. Topology of A2. Zariski topology is NOT the product topology.
In the product topology, the typical closed set is horizontal lines, vertical lines, and points.
In the Zariski Topology, the closed sets are unions of points and algebraic curves. Therefore, the
Zariski topology is finer than the product topology.
Example 2.9. A determinantal variety is an example of an algebraic set. Take:
Amn =m×nmatrices = linear maps : km → kn.
Look at the subset of matrices of rank ≤ N. This is an algebraic set; it is given by the subset
of all matrices such that all (N + 1) × (N + 1) submatrices have determinant 0. Recall that the
determinant is a polynomial in the entries of Amn.
Proposition 2.10. Any algebraic set is the union of a finite number of irreducible algebraic sets
(varieties).
Definition 2.11. An irreducible set is a set that cannot be written as the union of two smaller
closed subsets.
If a topological space is Hausdorff, the only irreducible sets are points, infinite closed sets are
not the union of finitely many irreducibles).
The assertion in Proposition 2.10 is true for any Noetherian topological space.
Definition 2.12. A ring is called Noetherian if its ideals satisfy one of the following:
(1) The Ascending Chain Condition (ACC). If
I1 ⊆ I2 ⊆ I3 ⊆ ··· .
is an ascending chain of ideals, it eventually stabilizes.
(2) Every ideal is finitely generated.
(3) Every nonempty set of ideals has a maximal element.
Remark 2.13. Affine space is Noetherian (as topological space) because k[x ,...,x ] (the ring of
1 n
n
polynomial functions on A is Noetherian as a ring.
Remark 2.14. If X ⊇ X ⊇ X ⊇··· is a descending sequence of algebraic sets, then I ⊆ I ⊆
1 2 3 1 2
I ⊆··· is an ascending sequence of algebraic sets, where I is the ideal of polynomials vanishing
3 k
on X .
k
Theorem 2.15 (Hilbert). If R is Noetherian, then R[x] is Noetherian.
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